@robjohn well maybe brute force is not the technical term, but there certainly isnt anything elegant about what I have in mind. In the past I wrote a (unoptimised) algorithm for visiting every face on a trinagle mesh, and was thinking of massaging the problem into that form...which I may do as a sanity check anyhow
Also I made an observation that I dont know how to formalise
If im on the right track maybe your silence is enough
If i am wildly off kilter perhaps silence is also good too
I want to explore my observation a little
I dont know how to say this without it sounding dumb, but I just noticed a relationship between the position of the verticies and wheather the line segments intersect inside the circle
Ok, i wont have time to solve it today (time is almost up before i turn into a grease monky again) but if i split the cicle into two sectors, how do I tell which sector a vertx is in?
My best guess is to use polar coordinates and define the sector by angle of the arc...
Ahhhh! A fly in my hotchoc...dammit maths
Got a way of expressing which sector the vertex is in, and now i must go. Thanks for the hint ;)
How do you write $ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2}$ in matrix form? Is it $\frac{1}{x^2}$ and $\frac{1}{y^2}$ on the diagonal and 0 else-where? I need this to calculate the volume form.
$g=\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ is the matrix representation of $g=ds^2=dx^2+dy^2$
so for $ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2}$ my best guess at the matrix representation is $\begin{bmatrix} \frac{1}{x^2} & 0 \\ 0 & \frac{1}{y^2} \end{bmatrix}$
ok but the notation is being mixed in weird ways. your det formula appears to be in notation that appears to be using x = (x^1, x^2) as a vector valued mapping, but up top with ds^2 you're using x as one coordinate
i'm playing "type mismatch" here instead of answering more directly because i think any confusion may be arising from this sort of thing and not deeper stuff
How do you write $ds^2=\frac{dx^2}{x^2}+\frac{dy^2}{y^2}$ in matrix form? Is it $\frac{1}{x^2}$ and $\frac{1}{y^2}$ on the diagonal and $0$ else-where? I need this to calculate the volume form.
$g=\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ is the matrix representation of $g=ds^2=dx^2+dy^2$
so f...
if your surface is parametrized as the image of some two-variable function r(x,y), then just using x and y, your computations show that the surface element dS oughta be 1/|xy| dx dy and you'd integrate some function f defined on the surface with respect to dS by integrating f(r(x,y)) 1/|xy| dx dy. i think this is just the usual thing from multivariable calculus
people often say something like area element or something when studying surfaces in 3 space although volume form is the general term for "hypersurfaces" in n-space
if you want to do a 'line integral' of a curve on the surface you'd integrate with respect to ds, not dS.
i may be confusing this by introducing new notation dS when it seems like you may already be consulting sources using varied notation that might not be mutually compatible with one another
i recommend picking one book and just following through how they do it. doesn't matter which one, just something that has a consistent universe of notation and discourse. some guy around here wrote one.
Any help with notation : I have a vector field $X: \mathbb{R} \times \mathbb{R}^d \to \mathbb{R}^d$ and the book says $X\in C(\mathbb{R};L)^N$ but has no reference for notation...
hi, why is it true that if $h \neq 0$ almost everywhere on some interval $(a,b)$, $-\infty \leq a < b \leq +\infty$, and $fh = 0$ almost everywhere on $(a,b)$, then $f = 0$ a.e on $(a,b)$?
couldn't we have say, $h = 1$ on $[0,1]$ and $0$ outside, and $f = 0$ on $[0,1]$ and $1$ outside, and then this would be false?
the underlying question i wanted to ask was, if I know that the fourier transform of a product of two functions is zero, and I know that at least one of the factors of the product is nonzero on a set of positive measure, can I conclude the other factor is zero a.e?
@porridgemathematics well, if $h=1$ on $[0,1]$ and $0$ outside, then it certainly isn't the case that $h\neq0$ almost everywhere for any interval properly containing $(0,1)$
and on the interval $(0,1)$, the conclusion holds, as it should
right, my underlying question was assume the fourier transform of a product of two functions in $L^2(a,b)$ is zero, and suppose one of the factors is not equal to $0$ in $L^2(a,b)$, is the other factor necessarily $0$ in $L^2(a,b)$?
Say I have two objects in a 3d euclidean space and I want to know if they intersect. Would it be a good idea to quickly negate intersection by bounding the objects in a convex hull and check if the two convex hulls dont intersect?
Because if the boundry volume does not intersect, the objects themselves surely dont
And if that is the case then, if I have two continuous functions $f$ and $g$ from metric space $X$ to metric space $Y$ with $E\subset X$ dense in $X$ and that $f(x)=g(x)$ for all $x\in E$
@Koro: you know that $f(e)=g(e)$ for all $e\in E$. So pick some $x\in X$ and $\epsilon\gt0$. Can you show that $d_Y(f(x),g(x))\le\epsilon$? (I had $E$ and $X$ reversed; now it's fixed)
I just opened this chat and saw your hint and before that I’d finished. I’ll explain my working now.
I assumed for contradiction that there is an $x\in X\setminus$ such that , $d_Y(f(x), g(x))\gt 0$. It follows by continuity of $f$ and $g$ at $x$ that given $\epsilon\gt 0$, there exists $\delta\gt 0$ such that $d_X(x,t)\lt \delta\implies d_Y(f(t),f(x))\lt \epsilon/2$ and $d_Y(g(x),g(t))\lt \epsilon/2$. Since $E$ is dense in $X$, we know that there is a $p\in E$ such that $d_X(x,p)\lt \delta$ whence it follows that $d_Y(f(x),f(p))\lt \epsilon/2$ and similarly for $g$.
Then by triangular inequality $d_Y(f(x),g(x))\le d_Y(f(x),f(p))+d_Y(f(p), g(x))$ and noticing that $f(p)=g(p)$ , we get a contradiction.
immediately proving our theorem.
Sir @robjohn, I suppose that you were hinting me to do this only. Is my proof correct?
rajorishi, you don't have to do any of those things but it sounds like you are talking about a common derivation of the quadratic formula via "completing the square"
if so, at a high level, the idea is to reduce solving an arbitrary quadratic to taking square roots
i.e., turning an arbitrary quadratic into $x^2 = \text{something}$, which is a solved problem
So if my epsilon is arbitrary and I have some $|x|\lt \epsilon $ where $\epsilon \gt 0$ is arbitrary then $|x|=0$. I have used this result. Is my proof correct now sirs @leslietownes @robjohn?
if you ask me to solve $x^2 = p$ the answers are $x = \pm \sqrt{p}$. i just take square roots
koro why do you get a contradiction? you are implicitly using the hypothesis that $d_Y(f(x),g(x)) > 0$ but it isn't explicit where. an evil instructor would say that you do not use that hypothesis at all.
think about the real case, suppose $f$ and $g$ are known to satisfy the hypothesis and also to be bounded by ten million. if epsilon is arbitrary, and i take epsilon = 20 billion, there's no contradiction whatsoever
ok, right. notice you didn't have that above. epsilon is arbitrary in applying the definition of the existence of the limit, but it isn't arbitrary in this proof, you're choosing that value of it
and the assumption that $d_y(f(x),g(x)) > 0$ is exactly what you use when you do that
if it wasn't assumed positive, it couldn't be plugged into the proof as an 'epsilon'
yeah, you've got it
The sequence approach I hinted at requires an understanding of the sequential consequences of density. Fix arbitrary $c$ in $X$. [Because $E$ is dense in $X$] there is a sequence $(e_n)$ in $E$ for which $e_n \to c$. Then $f(c) = \lim f(e_n) = \lim g(e_n) = g(c)$. This chain of equalities is justified by the continuity of $f$, $g$ (outer equalities) and because $e_n$ is in $E$ for all $n$ (inner equality; it means $f(e_n)=g(e_n)=0$ for all $n$).
it's sometimes fun if you prove something about metric spaces using open balls, to try to find a sequence proof, and vice versa. different people come at it different ways. sometimes it's easier to find one type of proof than the other, although which type may vary by person and problem
yes. if epsilon is arbitrary there is no way to decide whether there's anything wrong with $d_Y(f(x),g(x)) < \epsilon$. if $\epsilon = d_Y(f(x),g(x)) + 10$ for example, no contradiction
some people hate sequence proofs because they don't generalize to every conceivable space, and (particularly if the sequence has to be constructed in some complicated way) require reference to infinite sequences of things when (as in the open ball based proof) only a few parameters are often needed to get the job done.
while proving discontinuity of Dirichlet function (or is it also called Lebesgue function?). I mean the function on set of reals which assumes value 1 at rationals and 0 otherwise.
yeah, in some instances it's just easier to identify relevant sequences. the form of the function will suggest something specific that you don't have with just some arbitrary function on a metric space
as an exercise, come up with an open ball based proof for the dirichlet function :)
hartshorne's "geometry: euclid and beyond" has a very good discussion of them that would not offend you in its first chapters. it doesn't have a huge amount of exercises, but what's there is very good
hartshorne goes through the elements in some detail, pointing out where euclid cheats a little by using stuff that doesn't follow from his axioms or not handling all cases of a proof.
it's funny how every commentary or undergrad-level approach to the elements i have ever seen stops before the 3D stuff. i think by that point every author stares at that stuff and just thinks, if you want to do this, good luck.
yeah and hartshorne also recommends sections of euclid that are more worth reading. some sections introduce very fundamental ideas, others are just a grind. it seems hartshorne helps point this out
see if you can find my copy of the pre-springer edition that the postal service lost when i moved. it's probably in some used bookstore somewhere after some crook postal worker made a fortune off of selling it.
If I define $M$ as a mapping from $\mathbb{R}^d \to \mathbb{R}^d$, which is just matrix multiplication i.e $x\mapsto Ax$ for some invertible matrix $A$, then can If I define the pushforward of a (absolutely cont) probability measure $\mu$ on $\mathbb{R}^d$ as $M_{\#} \mu $ then what can we say about this guy ? is it a probability measure? hows it density relate to $\mu$
@leslietownes: I thought a lot about arbitrary epsilon. In fact, there is even a theorem for set of real numbers that if we have for some $x\in \mathbb R$, $0\lt x\lt \epsilon $ for any arbitrary $\epsilon \gt 0$, then $x=0$
So my question is back to square 1. Why do I have to specifically mention for what $\epsilon \gt 0$, I’ll have contradiction? $\epsilon \gt 0$ being arbitrary is itself an overkill.
koro your proof derives a contradiction from the fact that (after some inequalities) a deduction that $d_Y(f(x),g(x)) < \epsilon$. This will be a contradiction for some values of epsilon, but not others. You choose a value of epsilon that gives you a contradiction.
@koro if you ask me to derive a contradiction from the fact that some nonnegative function of a fixed $x$ is less than $\epsilon$, i can't do that without some idea of what $\epsilon$ is. if it's "arbitrary" it might well be a contradiction at all.
I understand for proving Cauchy sequence for example, if we have $ \sum _{n\gt N} x_n \lt \epsilon $ then here $N$ depends upon epsilon. We can’t use the aforementioned theorem. But in this case, x is fixed.
when people say epsilon is arbitrary in this context, they mean, because i have some positive number in mind, and the limit definition provides me with something for every positive epsilon, i can take epsilon to be that positive number. i don't have to worry that the limit definition won't provide me a delta in that context, because all i need to worry about is epsilon being positive. the limit definition will give me a delta as long as that happens.
ok, put that in everywhere i said it, same message.
if i make some assumption for the purpose of contradiction and deduce that 2 < epsilon, i don't know if i'm done unless i know why there's some reason why 2 can't be less than epsilon.
if i'm allowed to put epislon = 1/2 into the beginning (which i might be), then i get my contradiction. but if i'm doing that i should put that at the beginning of my proof. or say something like "since epsilon > 0 is arbitrary, and 1/2 is a positive number, we can put epsilon = 1/2 into the above analysis and deduce a contradiction" at the end of the proof.
@leslietownes: I would like to emphasize that limit definition is one thing as there we have delta depending upon epsilon . In my case above my $x$ is fixed and a fixed non negative quantity less than arbitrary positive values of epsilon implies that number has to be zero . I’m really interested in knowing how you can contradict this.
koro, in reference to your first proof, you just said "we get a contradiction" after deducing $d_Y(f(x),g(x) < \epsilon$. the arbitrary thing came later.
had you included a line, in that same set of lines, at the end saying "because epsilon > 0 is arbitrary, we get a contradiction" or something like that, i would have felt better about the proof.
it's also just clearer if you provide an example. unstated in that line of reasoning is "if $a \geq 0$ and $a < \epsilon$ for all $\epsilon$ then $a = 0$, and we assumed up above that $a$ was not zero." these are the jumps that many textbooks make in the later chapters but i would not make these jumps if the exercise is in chapter 1.
koro, that's the idea. there are two games, one to understand how to solve the problem, and two to include enough appropriately arranged detail to convince a reader, perhaps a very skeptical reader, that you haven't missed something. i think you won the first game and a lot of the above discussion was about the second.
and one or two lines, a change in order of wording, and things like that, can really affect game 2.
One final question for closure: Had I added this line "Since $\epsilon \gt 0$ is any arbitrarily chosen positive number, it follows that $d_Y(f(x),g(x))\epsilon \implies d_Y(f(x),g(x))\epsilon=0$ as metric $d_Y$ by definition can't be negative.", would my proof be correct/complete?
No no. I don't know why I'm typing so wrong today. $d_Y ((f(x),g(x))\lt \epsilon \implies d_Y (f(x),g(x))=0$ because $\epsilon \gt 0$ is any arbitrary positive number.
robjohn that's great. sometimes great horned owls perch in our neighborhood and fill the night with hoots but that seems to be a winter thing. no nesting.
while walking the dog in the park late one night, i thought i saw a child standing in the middle of the field. as i got closer, it took off. it was a massive great horned owl. it was so cool
Im in AZ so I would guess that what I saw was the north american great horned. it was probably because it was night, but when just standing out on the ground it looked so tall!
@robjohn that is a really great picture! i love the facial expression
I used this result: Suppose that $s$ is a non-negative real number such that for any arbitrary positive $\epsilon\gt 0$, we have $s\lt \epsilon$, then we have $s=0$. Proof: Suppose that $s\gt 0$ then since $\epsilon \gt 0$ is arbitray, let $\epsilon =\frac s2$ and by given hypothesis, we must have $s\lt \frac s2$ which means $2s-s\lt 0$ that is $s\lt 0$ which is a contradiction and therefore $s=0$.
@robjohn @leslietownes
I'm confused. I request your help in this @robjohn @leslietownes @copper.hat @TedShifrin
koro one reason there was a lot of static above is that it was not clear at the outset that you were using this result. from a classroom standpoint we also don't know whether you are 'allowed' to use this result (i.e., whether it was previously proved and citable, or whether you would be expected to include the proof you just gave as a sub-part of your argument). it's helpful to explicitly write out whatever path you are taking.
@Koro I am not really sure what it is that you are asking about? It is not easy to follow the thread of conversation to untangle exactly what you are trying to accomplish.
Yes. I accepted that earlier. I forgot to mention that in my first message explaining my working. And that's exactly what caused confusion. Now, my final question: with this idea included in my proof, my proof is complete. Right?
@copper.hat: I'm trying to prove that if $f,g$ are two continuous functions on metric space $X$ and let $E\subset X$ is dense subset of $X$ and given that $f(x)=g(x)$ for all $x\in E$ then prove that $f(x)=g(x)$ for all $x\in X$.
that makes sense considering \vspace is usually for the formatting of a whole document. hopefully people dont post things of that length as questions here!
im not sure i would have the patience to read it all
@robjohn yeah it makes more sense on main with the longer format. Are the same packages loaded on main as here? I really don't know very much about how mathjax works. I am procrastinating by reading mathjax documentation right now ha.
chat does not support $\LaTeX$, but the ChatJax bookmark uses the same header as main. Although, ChatJax still uses 2.7.1 and main uses 2.7.5. I should update the bookmark, but then people may see different things until they update.
Hopefully that will change, or they will update 2 if problems arise. Or there will be better documentation about what needs to be changed in terms of content.
I will create a bookmark using version 3 at some point and see what happens.
Hi everybody! Equations look nice when you're on the phone with the full version of the site. But it looks awful on the phone version. Why is that? I don't know what to do.
Here is the solution in a standard way:
We have,
$$\begin{align}a_1x_1^2+a_2x_2^2&=a_1\left(\frac{B-a_2x_2}{a_1}\right)^2+a_2x_2^2\\
&=\frac{B^2-2Ba_2x_2+a_2^2x_2^2}{a_1}+a_2x_2^2\\
&=\left(a_2+\frac{a_2^2}{a_1}\right)x_2^2-\left(\frac{2Ba_2}{a_1}\right)x_2+\frac{B^2}{a_1}\end{align}$$
This is th...
Hello I have a question. In this paper, page 978 eq. 32. For N\to\infty, \epsilon\to0. I know that the sum of the integral converges for N\to\infty. But the right hand side does not converge since \sum_{n\ge1}\mu(n) does not exist. This is all happening in one line and I cannot locate my mistake.
@Thorgott A proof can probaby be found in Edwards. I'm sure that the term on the LHS is 1. convergent 2. the same as the sum which runs over the trivial zeros.
